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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
RadiusThe length ofrope used is equalto the length oftthe radius of thecircle.What was the radius of the smallest circle? |
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| 2. |
A cone, a hemisphere and a cylinder stand on equal bases of radius R and have equalheights H. Find the ratio of their whole surface areas.0. |
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Answer» Volume of cone = (1/3)πr2hVolume of hemisphere = (2/3)πr3 Volume of cylinder =πr2h Given that cone, hemisphere and cylinder have equal base and same height. That is r = hVolume of cone : Volume of hemisphere : Volume of cylinder = (1/3)πr2h :(2/3)πr3:πr2h= (1/3)πr3:(2/3)πr3:πr3= (1/3) : (2/3) : 1= 1: 2: 3 |
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| 3. |
The radius of a circle is 9 cm. Find thelength of an arc of this circle whichcuts off a chord of length, equal tolength of radius. |
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Answer» Length of arc is 3pi cm |
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| 4. |
Height of a right circular cone is 8.4 cm and its base radius is 2.1cm. Find the radius of a spherewhose volume is equal to the volume of cone. |
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| 5. |
A cone, a hemisphere and a cylinder haveequal bases. If the heights of the cone andthe cylinder are equal to its commonradius, then the ratio between theirvolumes is:. |
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| 6. |
equal to length of radius.(3) Find in radians and degrees the angle subtended atthe centre of a circle by an arc whose length is15cms, ifthe radius of the circle is 25 cms. |
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| 7. |
NGLES8. In Fig. 6.54, O is a point in the interior of a triangleABC, OD I BC, OE LAC and OF LAB. Show that |
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| 8. |
(ii) 2p-1 = 23 |
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Answer» 2p-1= 232p= 24p= 24/2p= 12please like the solution 👍 ✔️ |
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| 9. |
23. If the equations Sx2+ (9+4p)r +2pfind the value of p.0 and 5x +90 are satisfied by the samevalueofx, |
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| 10. |
one by root 3 value |
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| 11. |
Find the HCF and LCM of 90 and 144 bymethod of prime factorization. |
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Answer» LCM bro 90 = 2 x3 x3 x5 = 2 x32x5144 = 2 x2 x2 x2 x3 x3 = 24x32HCF (90, 144) = 2 x32= 18LCM (90, 144) = 24x32x5 = 720 |
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| 12. |
ihI of 90 and 144 by the method of prime factorization |
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| 13. |
(a) 3n +7 = 25b) 2p - 1 = 23Solve: |
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Answer» a.3n =25-73n=18n=18/3n =6b.2p-1=23 2p=23+1 2p=24 p=24/2=12 |
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| 14. |
2. The lateral surface area of a cube is 100 m^2. The volume of the cube is(11) 1000 m^3(b) 100 m^3(c) 25 m^3d) 125 m^3 |
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Answer» lateral surface area=4a^2100=4a^2a^2=100/4=25a=5mhence volume =a^3=5^3=125m^3 |
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| 15. |
5,Houses in a row are numbered serially from49. Show that one value of c is such that sum ofnumbers of houses preceeding the markedhouse is equal to the sum of numbers of housessucceeding the marked house. Find the valueo |
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| 16. |
FIND THE HCF AND LCM op198 and 144 using primefactorization Method: |
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Answer» L cm =2×3×3×11×8=6×33×8=48×33=924; HCF =144[198-144]1=54]144-108]2=36[54-36]1=18[36-36[2/0 |
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| 17. |
Using prime factorization method, find the HCF and LCM of 6, 72 and 120. |
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Answer» 6=2*3 72=2*2*2*3*3 120=2*2*2*3*5 common factors=2*3=6(HCF) LCM=2*2*2*3*3*5=360(LCM) |
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| 18. |
Q4.Find the HCF and LCM of 198 and 144 using prime factorization method. |
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Answer» 144=2×2×2×2×3×3198=2×3×3×11 LCM(144,198)=Product of greatest power of each factor=2^4×3^2×11=1,584 HCF(144,198)=Product of lowest power of each common factor=2×3×3=18 |
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| 19. |
(95०० - 016 509 + 1)8100 (1) e®us—Digus+p ) :ownpa0 L’i =g102)1 |
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| 20. |
Find the HCF and LCMof 21, 15 and 12 using the prime factorizationmethod.9. |
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Answer» Explanation : 21 = 3×7 15 = 3×5 12 = 2×2×3 HCF means common factor of all the numbers so HCF = 3 LCM is the least common multiple of all 3 numbers so LCM = 3×7×5×2×2 = 420 Solution : HCF = 3 , LCM = 420 If you find this answer helpful then like it |
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| 21. |
Find the HCF and LCM of 6, 72 and 120, using the primefactorization method.4 |
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| 22. |
2 Write the following rational numbers in decimal form.127200237254(iv) |
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Answer» a)127/200=0.635b)25/99=0.2525..c)23/7=3.285d)4/5=0.8 will ypu plz explain me the method |
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| 23. |
(200995( 16( 1252. Write the following rational numbers in decimal form.(0) 127 (1) 25 (1) 23 (iv)Write the following rational numbers in form.(ii) 0.37(i) 0.6(iv) 15.89(iii) 3.17Let's recall.Let's recall.asional numbe |
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Answer» yuiy0v6vvy9c8xr8xr88ez7z7rxt8cy9v 3 (i) 6/10ii)37/100iii)317/100iv)1589/1000 |
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| 24. |
(12.20LUUL3. Convert each of the following into a decimal:2516(11) 156100100(vi) 3(vii) 2251000(iv) 3524(vii) 1720 |
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Answer» very easy questions you ask |
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| 25. |
Write the following decimal numbers in expanded form.0) 55.5(ii) 5.55 |
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| 26. |
25The houses of a row are numbered consecutively from 1 to 49. Show that there is a valueof x such that the sum of the numbers of the houses preceding the house numbered x isequal to the sum of the numbers of the houses following it. Find this value of x.4. |
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| 27. |
kem/ adAt ts根 Selling Price |
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Answer» Cp=2600profit=10%cost after getting profit=2600×10÷100=260selling price => 2600+260=2680 |
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| 28. |
UNDERSTAND NG ISC MA꺼EMATCn) The houses of a row are numbered from 1 to 49. Show that there is a value of n suchthat the sum of the numbers of the houses preceding the house numbered n is equalto the sum of the numbers of the houses following it. Find this value of n. |
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| 29. |
The houses of a row are numbered consecutively from 1 to 49. Show that there is a valuof x such that the sum of the numbers of the houses preceding the house numbered xequal to the sum of the numbers of the houses following it. Find this value ofx.Hint: S S4 S,] |
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Answer» thanks |
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| 30. |
The house in a row are numbered consecutively from 1 to 49. Show that there exists a value ofXsuch that sum of numbers of houses preceed ng the house numbered Xis equal to sum of thnumbers houses following x. Find value of x. |
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| 31. |
4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value4.of x such that the sum of the numbers of the houses preceding the house numbered xisqual to the sum of the numbers of the houses following it. Find this value of x.Hint:SSS49x |
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| 32. |
divide 150 into three parts such that the second number is five sixth the first and the third number is four fifth the second |
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| 33. |
21. The radii of two right circular cylinders are in the ratio 2:3 and their heights are in theratio 5:4. Calculate the ratio of their curved surface areas and also the ratio of theirvolumes |
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| 34. |
12.The radii of two right circular cylinders are in the ratio 2:3 & theirheight are in the ratio 5:4calculate the ratio of their curved surface areas& also the ratio of their v3, A cylindercal rond rollou |
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| 35. |
The ratio of circumferences of two circles is 5: 3. What is the ratio of dheir radi ?is the ratio of theirr |
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Answer» 5:3 is the answer. divide both circumferences A=Pi×r^2 Pi=constantC1:C2=R1:R2C1:C2= 5:3Therefore R1:R2=5:3Ans) The ratio between theradii of two circles is 5:3 |
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| 36. |
60. The radii of two right circular cylindersare in the ratio of 2 3 and their heightsare in the ratio of 5 4. Calculate the ratioof their curved surface areas and ratio of[CBSE 2012]their volumes. 5 g |
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| 37. |
Find the L.C.M. of 84, 90 and 120 by prime factorization method. |
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| 38. |
Find HCF and LCM of 56 and 112 by prime factorization method. |
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| 39. |
el: Find the HCF of 72 and 90 by prime factorization method. |
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Answer» The given numbers are72, 108 and 180 We have: 272236218393312722362183933121082543273933121082543273933121802903453155512180290345315551Now, 72=2 × 2 × 2 × 3 × 3 = 23× 32108 = 2 × 2 × 3 × 3 × 3 = 22× 33180 = 2 × 2 × 3 ×3 × 5 = 22× 32× 5∴ HCF =22×32=36 |
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| 40. |
le 2 Find the H.C.F. of 1260 and 2376 by prime factorization method. |
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Answer» thanks for helping me.i will explain it to my son |
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| 41. |
35. The height of a right circular cylinder is 10.5m. Three times the sum of the areas of its twocircular faces is twice the area of the curved surface find the volume of the cylinder |
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| 42. |
Verify thatkつcays = (arty) (χ.xytta_ )。 |
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Answer» R.H.S =>(x+y) ( x^2- xy + y^2)=> x^3- x^2y + xy^2+ x^2y - xy^2+ y^3 {On multiplying x3+ y3with(x+y) ( x^2 - xy + y^2)}=> [x^3+ y^3] + ( -x^2y+ x^2y) + (xy^2- xy^2)=> x^3+ y^3Since R.H.S = L.H.S,That is x^3+ y^3= x^3+ y^3Hence, verified that x3+ y3=(x+y) ( x2- xy + y2) |
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| 43. |
What is the concentration of sugar (C12H201) in mol L if its 20 g are dissolved inenough water to make a final volume up to 2L? |
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| 44. |
50 circular plates, each of radius 7 cm and thickness-cm, \are placed one above another to form a solid right circularcylinder. Find the total surface area and the volume of thecylinder so formed.2. |
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Answer» Since the radius of the cylinder is same asthe radius of the circular plate.So, the radius of the base of the cylinder = 7 cmHeight of the cylinder is same as the thickness of 50 circular plates.Thickness of 1 plate = 1/2 cmThen, thickness of 50 plates = 50*1/2= 25 cmHeight of the cylinder = 25 cmTotal surface area of cylinder = 2πr(r+h)= 2*22/7*7(7+25)= 44*32Total surface area of the cylinder = 1408 sq cmVolume of the cylinder =πr²h= 22/7*7*7*25= 3850 cu cmVolume of the cylinder is 3850 cu cm |
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| 45. |
6.d the mean of first five multiples of 3. |
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| 46. |
37. The sum of the radius of base andheight of a solid right circular cylinder is37 cm. If the total surface area of the solidcylinder is 1628 cm2, find the volume ofthe cylinder. (UseICBSE 2016] |
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| 47. |
A Caん.tand 60 krn İn / hん30'min. HowGo lamฮ่レvolllieed 1 |
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| 48. |
replacing n by 2 x 12-1-Hence, required ratio (69 + 8):(161 +15) 7:16.Ifthere are (2n + 1)terms in AP, then prove that the ratio of the sum oEXAMELE 31sum of even terms is (n+1):n.NCEIls tho first term and common difference respectivelv |
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| 49. |
46. How are the words in the document counted? |
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Answer» To see theword countin yourdocument, look at the status bar at the bottom of theWordwindow. Wordcan insert theword countinto yourdocumentand update that information as often as you want. Click in yourdocumentwhere you want theword countto appear. In the Field names list, click NumWords, and then click OK. |
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| 50. |
Q.16. Find the value of m so that 2x-1 be a factor of 8x* +4x°-16x2 +10x+m. |
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